Tìm GTLN hoặc GTNN của bt sau E=x^2-xy+3y^2-2x-10y+20

Đáp án:

$\min E = 8\Leftrightarrow x = y= 2 $

Giải thích các bước giải:

$\begin{array}{l}\quad E = x^2-xy+3y^2-2x-10y+20\\ \to 4E = 4x^2 – 4xy + 12y^2 – 8x – 40y +80\\ \to 4E = (4x^2 – 4xy + y^2 -8x – 4y + 4) + (11y^2 -44y+76)\\ \to 4E = (2x -y – 2)^2 + 11(y^2 -4y + 4) + 32\\ \to 4E = (2x – y – 2)^2 + 11(y – 2)^2 + 32\\ \to E = \dfrac14(2x – y – 2)^2 + \dfrac{11}{4}(y-2)^2 + 8\\ \text{Ta có:}\\ \quad \begin{cases}(2x – y -2)^2 \geq 0\quad \forall x;y\\(y-2)^2 \geq 0\quad \forall x\end{cases}\\ \to \dfrac14(2x – y – 2)^2 + \dfrac{11}{4}(y-2)^2 +8 \geq 8\\ \to E \geq 8\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \begin{cases}2x – y – 2= 0\\y – 2=0\end{cases}\Leftrightarrow x =y =2\\ Vậy\,\,\min E = 8\Leftrightarrow x = y= 2 \end{array}$